$\frac{1}{a^2 - 1} = \frac{1}{(a+1)(a-1)} = \frac{1}{2} \left (\frac{1}{a-1} - \frac{1}{a+1} \right )$

So, I can write the given sum as :

$\frac{1}{2} \left [ \left (\frac{1}{2-1} - \frac{1}{2+1} \right ) + \left (\frac{1}{4-1} - \frac{1}{4+1} \right ) + \left (\frac{1}{6-1} - \frac{1}{6+1} \right ) +……. + \left (\frac{1}{40-1} - \frac{1}{40+1} \right ) \right ]$

$\frac{1}{2} \left [ \left (\frac{1}{1} - \frac{1}{3} \right ) + \left (\frac{1}{3} - \frac{1}{5} \right ) + \left (\frac{1}{5} - \frac{1}{7} \right ) + \frac{1}{7} – ……. + \left (\frac{1}{39} - \frac{1}{41} \right ) \right ]$

$\frac{1}{2} \left ( \frac{1}{1} – \frac{1}{41} \right)$ $(\because \frac{1}{a+1} = \frac{1}{(a+2)-1})$

$\frac{20}{41} $